Differentiate. $$ f(t)=\frac{2 t}{9+t^{2}} $$ $$ f^{\prime}(t)=\square $$

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UpStudy AI · Accepted Answer

1. Recall the quotient rule for differentiation: if 
   $$
   f(t)=\frac{g(t)}{h(t)},
   $$
   then
   $$
   f'(t)=\frac{g'(t)h(t)-g(t)h'(t)}{[h(t)]^2}.
   $$

2. Identify $$ g(t) $$ and $$ h(t) $$:
   $$
   g(t)=2t \quad \text{and} \quad h(t)=9+t^2.
   $$

3. Differentiate $$ g(t) $$ and $$ h(t) $$:
   $$
   g'(t)=2,
   $$
   $$
   h'(t)=2t.
   $$

4. Substitute into the quotient rule:
   $$
   f'(t)=\frac{2(9+t^2)-2t(2t)}{(9+t^2)^2}.
   $$

5. Simplify the numerator:
   $$
   2(9+t^2)=18+2t^2,
   $$
   and 
   $$
   2t(2t)=4t^2.
   $$
   Thus,
   $$
   18+2t^2-4t^2 = 18-2t^2.
   $$

6. Factor the common factor in the numerator:
   $$
   18-2t^2=2(9-t^2).
   $$

7. The derivative is:
   $$
   f'(t)=\frac{2(9-t^2)}{(9+t^2)^2}.
   $$
$$ f^{\prime}(t)=\frac{2(9-t^2)}{(9+t^2)^2} $$

Differentiate.

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